Question

if $\sf A = \left[ \begin{array}{cc} 1 & \sf tan\:x\\\sf -tan\: x & \sf 1 \end{array}\right]$

then find $\rm A^T A^{-1}$​

Answer 1

$\LARGE\mathcal{ \:\;\:\:\:\:\:\:\;\underline{\underline{ANSWER}}}$

$\rm$

$\sf |A| = \left|\begin{array}{cc}\sf 1 & \sf tan\:x\\\sf -tan\: x & \sf 1 \end{array}\right|$

$\sf = (1)(1) \: - \: (-tan\:x)(tan\:x)$

$\sf \implies 1+tan^2 \: x \neq 0$

$\rm \therefore$ A is invertible.

let $\rm C_{\sf ij}$ be the cofactor of $\sf a_{ij}$ in $\sf A = [a_{ij}]$

$\sf \therefore C_{11} = (-1)^{1+1} 1= 1$

$\sf\:\:\:\: C_{12} = (-1)^{1+2} (-tan\: x) = tan\: x$

$\sf \:\:\:\:C_{21} = (-1)^{2+1} tan \: x = -tan \: x$

$\sf \:\:\:\;C_{22} = (-1)^{2+2} \cdot 1 = 1$

$\rm$

$\sf \therefore \rm{adj} \: \sf A = \left[ \begin{array}{cc} \sf 1 & \sf tan\:x\\\sf -tan\: x & \sf 1\end{array}\right]^T$

$\sf = \left[ \begin{array}{cc} \sf 1 & \sf -tan\:x\\\sf tan\: x & \sf 1\end{array}\right]$

Now, $\sf A^{-1} = \dfrac{1}{(1+tan^2\:x)} \left[ \begin{array}{cc} \sf 1 & \sf -tan\:x\\\sf tan\: x & \sf 1\end{array}\right]$

$\sf = \left[ \begin{array}{cc}\sf \dfrac{1}{1+tan^2\:x} &\sf \dfrac{-tan\:x}{1+tan^2\:x} \\\\\sf \dfrac{tan\:x}{1+tan^2\:x} &\sf \dfrac{1}{1+tan^2\:x} \end{array}\right]$

$\sf\therefore A^T A^{-1} = \left[ \begin{array}{cc} \sf 1 & \sf -tan\:x\\\\\sf tan\: x & \sf 1\end{array}\right]\left[ \begin{array}{cc}\sf \dfrac{1}{1+tan^2\:x} &\sf \dfrac{-tan\:x}{1+tan^2\:x} \\\\\sf \dfrac{tan\:x}{1+tan^2\:x} &\sf \dfrac{1}{1+tan^2\:x} \end{array}\right]$

$\sf \left[\begin{array}{cc}\sf \dfrac{1-tan^2\:x}{1+tan^2\:x} &\sf \dfrac{-2\:tan\:x}{1+tan^2\:x} \\\\\sf \dfrac{2\:tan\:x}{1+tan^2\:x} &\sf \dfrac{1-tan^2\:x}{1+tan^2\:x} \end{array}\right]$

$\boxed{\boxed{\underline{= \left[\begin{array}{cc}\sf cos \: 2x &\sf -sin\: 2x\\\sf sin\: 2x &\sf cos \: 2x\end{array}\right]}}}$

Additional Information:-

determinant of $\sf A^TA^{-1}$

$\sf =\left|\begin{array}{cc}\sf cos \: 2x &\sf -sin\: 2x\\\sf sin\: 2x &\sf cos \: 2x\end{array}\right|$

$\sf = (cos\:2x)(cos\:2x)-(-sin\:2x)(sin\:2x)$

$\sf = cos^2\: 2x + sin^2\: 2x$

use the double angle formula

$\sf = (cos^2x-sin^2x)^2 +(2\: sinx\: cosx)^2$

$\sf=cos^4x-2sin^2x\:cos^2x+sin^4x+4sin^2x\:cos^4x$

$\sf = cos^4 x +2sin^2x\:cos^2x + sin^4x$

$\sf = (cos^2 x+sin^2 x)^2$

$\sf = 1^2 = 1$

$\boxed{\sf \implies \left|A^TA^{-1}\right| = 1}$

Answer 2

$\large{\text{\underline{\rm{Given :}}}}$

$\sf{A}=\left[\begin{array}{cc}1&\sf{tanx}\\\sf{-tanx}&5\end{array}\right]$

$\large{\text{\underline{\rm{To Find :}}}}$

$\bf A^TA^{-1}$

$\large{\text{\underline{\rm{Solution :}}}}$

$\bf A= \left[\begin{array}{cc}1&\sf tanx\\\sf -tanx&1\end{array}\right]$

$\bf A^T = \left[\begin{array}{ccc}1&\sf-tanx \\\sf tanx&1\end{array}\right]$

$\bf A^{-1}=\frac{1}{|A|}adjA$

$\bf{|A|}=(1)(1)-((\sf -tanx)(tanx))\implies(1)-(-tan^2x)$

$\bf{|A|}=\sf1+tan^2x \neq 0$

$\because \bf{A} \sf \:\:is\:\: invertible$

$\text{Let's have C_{ij} as a co-factor of a_{ij} }$

$C_{11}=(-1)^{1+1}\:(1)\implies(-1)^2(1)\implies 1$

$C_{12} = (-1)^{1+2} \:\: \sf{(tanx)}\implies (-1)^3\:\sf{(tanx)}\implies \sf -tanx$

$C_{21}=(-1)^{2+1}(\sf -tanx)\implies(-1)(\sf -tanx) \implies \sf tanx$

$C_{22}= (-1)^{2+2}(1)\implies (-1)^4(1)\implies 1$

$\bf{ adjA}=\left[\begin{array}{ccc}1&\sf-tanx\\\sf tanx&1\end{array}\right]$

$\bf A^{-1}=(\frac{1}{\sf 1+tan^2x})\left[\begin{array}{cc}1&\sf-tanx\\\sf tanx&1\end{array}\right]$

$\bf A^{-1}=\large{\left[\begin{array}{ccc}\sf\frac{1}{1+tan^2x}&\frac{\sf -tanx}{\sf 1+tan^2x}\\\frac{\sf tanx}{\sf 1+tan^2x}&\frac{1}{\sf 1+tan^2x}\end{array}\right] }$

$\bf{ A^TA^{-1}}=\left[\begin{array}{ccc}1&\sf -tanx\\\sf tanx &1\end{array}\right]\times\large{\left[\begin{array}{ccc}\sf\frac{1}{1+tan^2x}&\frac{\sf -tanx}{\sf 1+tan^2x}\\\frac{\sf tanx}{\sf 1+tan^2x}&\frac{1}{\sf 1+tan^2x}\end{array}\right] }$

$\bf{ A^TA^{-1}}=\large{\left[\begin{array}{ccc}(\frac{1}{\sf 1+tan^2x}-\frac{\sf tan^2x}{\sf 1+tan^2x})&(\frac{\sf -tanx}{\sf 1+tan^2x}-\frac{\sf tanx}{\sf 1+tan^2x})\\(\frac{\sf tanx}{\sf 1+tan^2x}+\frac{\sf tanx}{\sf 1+tan^2x}) &(\frac{1}{\sf 1+tan^2x}-\frac{\sf tan^2x}{\sf 1+tan^2x})\end{array}\right]}$

$\bf{ A^TA^{-1}}=\large {\left[\begin{array}{ccc}\frac{\sf 1-tan^2x}{\sf1+tan^2x}&\frac{\sf-2\:tanx}{\sf 1+tan^2x}\\\frac{\sf 2\:tanx}{\sf 1+tan^2x}&\frac{\sf 1-tan^2x}{\sf1+tan^2x}\end{array}\right] }$

$\bf{ A^TA^{-1}}=\large{\left[\begin{array}{ccc}\frac{1-\frac{\sf sin^2x}{\sf cos^2x}}{1+\frac{\sf sin^2x}{\sf cos^2x}}&\frac{-2\times \frac{\sf sinx}{\sf cosx }}{1+\frac{\sf sin^2x}{\sf cos^2x}}\\\frac{2\times \frac{\sf sinx}{\sf cosx }}{1+\frac{\sf sin^2x}{\sf cos^2x}} &\frac{1-\frac{\sf sin^2x}{\sf cos^2x}}{1+\frac{\sf sin^2x}{\sf cos^2x}}\end{array}\right]}$

$\bf{ A^TA^{-1}}=\large{\left[\begin{array}{ccc}\frac{\frac{\sf cos^2x-sin^2x}{\sf cos^2x}}{\frac{\sf cos^2x+sin^2x}{\sf cos^2x}}&\frac{-2 \times \frac{\sf sinx}{\sf cosx}}{\frac{\sf cos^2x+sin^2x}{\sf cos^2x}}\\\frac{2 \times \frac{\sf sinx}{\sf cosx}}{\frac{\sf cos^2x+sin^2x}{\sf cos^2x}}&\frac{\frac{\sf cos^2x-sin^2x}{\sf cos^2x}}{\frac{\sf cos^2x+sin^2x}{\sf cos^2x}}\end{array}\right]}$

$\bf{ A^TA^{-1}} = \large{ \left[\begin{array}{ccc}\frac{\sf cos^2x-sin^2x}{\sf cos^2x+sin^2x}&\frac{-2\times \frac{\sf sinx}{\sf cosx }}{\frac {1}{\sf cos^2x}}\\\frac{2\times \frac{\sf sinx}{\sf cosx }}{\frac {1}{\sf cos^2x}}&\frac{\sf cos^2x-sin^2x}{\sf cos^2x+sin^2x}\end{array}\right]}$

$\color{deeppink}{\sf (\:\:\:\: \because sin^2+cos^2=1\:)}$

$\bf{ A^TA^{-1}} = \left[\begin{array}{ccc}\frac{\sf cos\:2x}{1}&(-2\frac{\sf sinx}{\sf cosx}\times \sf cos^2x)\\(2\frac{\sf sinx}{\sf cosx}\times \sf cos^2x)&\frac {\sf cos\:2x}{1}\end{array}\right]$

$\color{deeppink}{\sf (\:\:\: \because cos^2x-sin^2x=cos2x)$

$\bf{ A^TA^{-1}} = \left[\begin{array}{ccc}\sf cos\:2x&\sf -2\:sinx\:cosx\\\sf 2\:sinx\:cosx&\sf cos\:2x\end{array}\right]$

${\pink{\sf (\:\: \because\:\:2\:sinx\:cosx=sin\:2x)}}$

$\bf{ A^TA^{-1}} = \left[\begin{array}{ccc}\sf cos\:2x&\sf -sin\:2x \\\sf sin\:2x&\sf cos\:2x\end{array}\right]$